/*
 Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
 

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
 

Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
 

Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
 

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6
 */
package com.yuan.algorithms.acm201505;

import java.util.Scanner;

public class HD1003MaxSum {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int n = sc.nextInt();
			while (n-- > 0) {
				int t = sc.nextInt();
				int sum = 0, max = -0xfffffff;
				int start = 1, end = 1, temp = 1,cas = 1;
				for (int i = 1; i <= t; i++) {
					int k = sc.nextInt();
					sum += k;
					if (sum > max) {
						max = sum;
						start = temp;
						end = i;
					}
					if (sum < 0) {
						temp = i + 1;
						sum = 0;
					}
				}
				if (cas!=1) {
					System.out.println();
				}
				System.out.println("Case "+cas+":");
				System.out.println(max+" "+start+" "+end);
			}
		}
	}

}
